anyone?

Is this an old putnam problem?

This is a preparation question for the qualification to the math olympics, I just can't figure this one out, you aren't supposed to use a calculator, only pen and paper.

it's close to 10% with 3rd, 13th etc starting with 4 (at least for the initial segment of the sequence)
just wondering when the sequence 1024, 1024^2, 1024^3, starts to run over to a starting 2

This post was edited by brmv on Nov 25 2013 01:13am

I was going to say... probably close to a third of 605.

Since the 606 digit number (largest one) starts with a 1, there are 605 possibilities for numbers starting with a 4. 4x is missed, as is 4xx...

Tough one.

2010/10=201 and 605/3=201.667 are pretty close to each other

we know that when the number of digits increases it will always have a 1 first, so be something between 1.000... and 1.999...
but when the number is 1.25 or more the 4 will be missed and one has to account for the 4 then coming at the next digit: 1.25, 2.5, 5.0.10.0, 20.0, 40.0
and we know that it takes not more than 10 steps to increase the number of digits by 3

taking all together it seems that the 201 is rather an upper limit and the exact number will be lower

Can someone tell me how 2 times anything will give you an odd number?

195?

there is a difference of 3 in the powers, 13th and 10th
occurring 20 times, 60/10=6
201-6
however, this was done more visually with aid of computer :/

This post was edited by saber_x3 on Nov 25 2013 02:23am

try 16, multiply with 2 gives 32 and 3 is an odd number or isn't it?
seems you have misread the problem

the answer is approximately 2009/10

the beauty of power of 2's is that is virtually repeats...therefore you can conclude:

2^1=2
2^2 = 4
2^3 =8
2^4 = 16
2^5 = 32
2^6 = 64
2^7= 128
2^8=256
2^9=512
2^10=1024
2^11=2048
2^12=4096
2^13=8192

this will forever repeat....so 2^2, 2^12. 2^22, 2^32, 2^42, 2^52, etc etc etc -- eventually it will turn to something like 2^155, but it will only happen once ever 10 digit increases...this is a quick and dirty way of estimating that approximately 200 numbers will have a 4-handle.

This post was edited by mungoflago on Nov 25 2013 02:12pm